JEE Advanced 2024 Paper 2 · Q17 · Spring-Block
[PARAGRAPH II] Two particles, 1 and 2, each of mass $m$, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at $x_0$, are oscillating with amplitude $a$ and angular frequency $\omega$. Thus, their positions at time $t$ are given by $x_1(t) = (x_0 + d) + a\sin\omega t$ and $x_2(t) = (x_0 - d) - a\sin\omega t$, respectively, where $d > 2a$. Particle 3 of mass $m$ moves towards this system with speed $u_0 = a\omega/2$, and undergoes instantaneous elastic collision with particle 2, at time $t_0$. Finally, particles 1 and 2 acquire a center of mass speed $v_{\text{cm}}$ and oscillate with amplitude $b$ and the same angular frequency $\omega$.
If the collision occurs at time $t_0 = \pi/(2\omega)$, then the value of $4b^2/a^2$ will be _____.
Reveal answer + step-by-step solution
Correct answer:4.25
Solution
At $t_0 = \pi/(2\omega)$: $\sin\omega t_0 = 1$, $\cos\omega t_0 = 0$. So velocities just before collision: $\dot x_1 = a\omega\cos\omega t_0 = 0$, $\dot x_2 = -a\omega\cos\omega t_0 = 0$. Both particles momentarily at rest at extreme positions: $x_1 = x_0 + d + a$, $x_2 = x_0 - d - a$.
Particle 3 hits particle 2 with $v_3 = +a\omega/2$, $v_2 = 0$. Elastic, equal masses ⇒ velocities exchange: $v_3' = 0$, $v_2' = a\omega/2$.
After collision, the (1,2) system: position of particle 1 still at extreme ($x_0 + d + a$), velocity 0; particle 2 at $x_0 - d - a$ with velocity $a\omega/2$.
Centre of mass: $X_{\text{cm}} = (x_1 + x_2)/2 = x_0 + (a - a)/2 = x_0$. $v_{\text{cm}} = (0 + a\omega/2)/2 = a\omega/4$.
In the COM frame, relative coordinate $y = x_1 - x_2 - 2d_{\text{eq}}$... cleanest to use energy. Total KE in COM frame just after collision: Velocities relative to COM: $u_1 = 0 - a\omega/4 = -a\omega/4$; $u_2 = a\omega/2 - a\omega/4 = a\omega/4$. Symmetric. Reduced mass $\mu = m/2$. Relative speed $\dot r = u_2 - u_1 = a\omega/2$.
Separation between particles: $x_1 - x_2 = (x_0 + d + a) - (x_0 - d - a) = 2d + 2a$. Equilibrium separation is $2d$. So spring extension $=…[truncated]
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